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I am suggested the following formula and I can't find some reference on this expression :

$$\text{Va}r(X/Y)=E(X^2/Y^2)-E^2(X/Y)=\text{Cov}(X^2,1/Y^2)+E(X^2)E(1/Y^2)-(\text{Cov}(X,1/Y)+E(X)E(1/Y))^2$$

I tried to demonstrate it without success.

EDIT:

From the first answer below, it seems that we should have rather the expression :

$\text{Var}(X/Y)={\rm cov}(X^{2}, 1/Y^{2}) + [{\rm var}(X) + E(X)^2] \cdot[{\rm var}(1/Y) + E(1/Y)^2] - [ {\rm cov}(X,1/Y) + E(X)E(1/Y) ]^{2}$

Do you agree with the validity of this expression ?

By the way : if I sample a random variable following a distribution and after that, I compute the variance of the inverse of sample (1/X) : will I have the same result if I sample directly the random (1/X) and compute after the variance of the sample ? I say that since I don't know a priori the distribution of 1/X even if I know the distribution of X.

What do you think about this key point ?

youpilat13
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1 Answers1

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The expression is not generally valid. Suppose $X=Y$ and $E[1/Y]$ does not exist. Then the left-hand side is 0 and the right-hand side is undefined.

If all the necessary expectations exist, take $Z=1/Y$ and compute $\mathrm{Var}[XZ]$ as described in this answer.

Thomas Lumley
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    The necessary conditions are essentially that $E[X^2]$ and $E[1/Y^2]$ (i.e. $E[Z^2]$) are finite – Henry Sep 17 '21 at 08:17